Is the Exact Differential Equation Solution Valid for Another Equation?

[sunnyceej]

Homework Statement

Consider the equation (y^2+2xy)dx-x^2dy=0 (a) Show that this equation is not exact. b) Show that multiplying both sides of the equation by y^-2 yields a new equation that is exact. C) use the solution of the resulting exact equation to solve the origional equation. d) Were any solutions lost in the process?

Homework Equations

How do you show that the solution in part b is a solution to the original?

The Attempt at a Solution

answer to part b: (x+x^2y^-1=c)
I tried (y^2+2xy)dx - x^2dy=(x+x^2y^-1) and integrated both sides to get
xy^2 + x^2y - x^2y = xy+x^2 ln y

I tried solving (x+x^2y^-1=c) for x: getting x=-x^2y^-1 +c and plugging in into the original equation : (y^2+2(-x^2y^-1 +c)y)dx - (-x^2y^-1 +c)^2dy=0 and that just got me a mess. I'm not sure what else to try.
I'm not sure how to do part d, either, but I figured I needed part c first.

 

[tiny-tim]

welcome to pf!

hi sunnyceej! welcome to pf!

(try using the X² button just above the Reply box )

answer to part b: (x+x^2y^-1=c)

yes

but after that, i don't understand what you're doing

just rewrite x + x²/y = C in the form y = … ​

 

[sunnyceej]

Thanks! I got y=x^2/(c-x)

Every time I use the x² button at the top I get SUP/SUP

I thought I also had to plug it back into the original to verify, but I found out solving for y is good! :)

 

[SammyS]

Thanks! I got y=x^2/(c-x)

Every time I use the x² button at the top I get SUP/SUP

As a result, your x[ SUP]2[ /SUP] looks like x² as it should.